1. The list below shows the number of text messages that five students sent from
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The difference between the rate of change of B and the rate of change of A is 15.
<h3>How to get the difference in the rates of change?</h3>
First, we need to get the two rates of change.
We have two linear functions.
A: y = 35*x
For function A the rate of change is the slope, wich is 35.
Function B is graphed.
Remember that if a line passes through two points (x₁, y₁) and (x₂,y₂) then the slope is:
In the graph, we can see that line B passes through (0, 0) and (1, 50), then the slope is:
Then the rate of change of B is 50.
The difference between the rates of change is:
diff = 50 - 35 = 15
If you want to learn more about rates of change:
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Answer:
A) C1 = 0.00187 m = 0.187 cm, C2 = 0.0062 m = 0.62 cm
B) A sample of how the graph looks like is attached below ( periodic sine wave )
C) w = is when the amplitude of the forced response is maximum
Step-by-step explanation:
Given data :
mass = 5kg
length of spring = 10 cm = 0.1 m
f(t) = 10sin(t) N
viscous force = 2 N
speed of mass = 4 cm/s = 0.04 m/s
initial velocity = 3 cm/s = 0.03 m/s
Formulating initial value problem
y = viscous force / speed = 2 N / 0.04 m/s = 50 N sec/m
spring constant = mg/ Length of spring = (5 * 9.8) / 0.1 = 490 N/m
f(t) = 10sin(t/2) N
using the initial conditions of u(0) = 0 m and u"(0) = 0.03 m/s to express the equation of motion
the equation of motion = 5u" + 50u' + 490u = 10sin(t/2)
A) finding the solution of the initial value
attached below is the solution and
B) attached is a periodic sine wave replica of how the grapgh of the steady state solution looks like
C attached below
